+0

0
53
3
+56

Find constants A and B such that

$$\frac{x+7}{x^2 -x-2}=\frac{A}{x-2} + \frac{B}{x+1}$$

for all x such that $$x \ne -1$$ and $$x \ne 2$$. Give your answer as the ordered pair (A,B).

Thank you

May 8, 2021

#1
0

If you let x go infinity, then you get 0 = A + B.

If you let x = 0, then you get -7/2 = A/(-2) + B.

The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3).

May 8, 2021
#2
+2

Note that x^2-x-2   factors to   x-2   *   x+1

Multiply the  equation by   (x-2)(x+1)

x+7      =   a(x+1)   + b (x-2)

x+7   = ax + a + bx - 2b             equat the 'x' and 'non-x' components to get the following two equations

x = x (a+b)             and           7 = a - 2b

1 =( a+b )

a = 1-b       sub this into the red equation     7 = 1-b    - 2b       to find   b = -2    then a = 3

May 8, 2021
#3
+56
0

Thank you!

shananigans  May 8, 2021